{"id":248754,"date":"2025-07-08T17:59:52","date_gmt":"2025-07-08T17:59:52","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=248754"},"modified":"2025-07-08T17:59:54","modified_gmt":"2025-07-08T17:59:54","slug":"for-a-projectile-the-ratio-of-maximum-height-reached-to-the-square-of-flight","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/08\/for-a-projectile-the-ratio-of-maximum-height-reached-to-the-square-of-flight\/","title":{"rendered":"For a projectile, the ratio of maximum height reached to the square of flight"},"content":{"rendered":"\n<pre id=\"preorder-ask-header-text\" class=\"wp-block-preformatted\">. For a projectile, the ratio of maximum height reached to the square of flight<\/pre>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/image-317.png\" alt=\"\" class=\"wp-image-248755\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The correct answer is&nbsp;<strong>1) 5:4<\/strong>.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Here is a detailed explanation of the solution.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">This problem requires understanding the fundamental equations of projectile motion. We need to find the ratio of the maximum height a projectile reaches to the square of its total time of flight. Let&#8217;s denote the initial velocity of the projectile as &#8216;u&#8217;, the angle of projection with the horizontal as &#8216;\u03b8&#8217;, and the acceleration due to gravity as &#8216;g&#8217;.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">First, we express the formula for the maximum height (H_max). The maximum height is achieved when the vertical component of the projectile&#8217;s velocity becomes zero. Using the kinematic equation v_y\u00b2 = u_y\u00b2 + 2as, where v_y = 0, u_y = u sin(\u03b8), and a = -g, we get:<br>0 = (u sin(\u03b8))\u00b2 &#8211; 2gH_max<br>This rearranges to give the formula for maximum height:<br><strong>H_max = (u\u00b2 sin\u00b2\u03b8) \/ (2g)<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Next, we determine the formula for the total time of flight (T). The time of flight is the total duration the projectile spends in the air. It is twice the time it takes to reach the maximum height. The time to reach the peak (t_peak) can be found using v_y = u_y + at:<br>0 = u sin(\u03b8) &#8211; gt_peak<br>So, t_peak = (u sin\u03b8) \/ g.<br>The total time of flight is T = 2 * t_peak, which gives:<br><strong>T = (2u sin\u03b8) \/ g<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The question asks for the ratio of the maximum height to the square of the flight time (T\u00b2). First, let&#8217;s find the expression for T\u00b2:<br><strong>T\u00b2 = [(2u sin\u03b8) \/ g]\u00b2 = (4u\u00b2 sin\u00b2\u03b8) \/ g\u00b2<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Now, we can form the required ratio, H_max \/ T\u00b2:<br>Ratio = H_max \/ T\u00b2 = [ (u\u00b2 sin\u00b2\u03b8) \/ (2g) ] \/ [ (4u\u00b2 sin\u00b2\u03b8) \/ g\u00b2 ]<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">To simplify this complex fraction, we can multiply by the reciprocal of the denominator:<br>Ratio = (u\u00b2 sin\u00b2\u03b8 \/ 2g) * (g\u00b2 \/ 4u\u00b2 sin\u00b2\u03b8)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">We can now cancel out the common terms. The terms u\u00b2 and sin\u00b2\u03b8 appear in both the numerator and the denominator, so they cancel. One &#8216;g&#8217; from the numerator cancels with the &#8216;g&#8217; in the denominator:<br>Ratio = g \/ (2 * 4) = g \/ 8<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The problem specifies to use g = 10 m\/s\u00b2. Substituting this value into our simplified expression:<br>Ratio = 10 \/ 8<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Finally, simplifying this fraction gives:<br>Ratio = 5 \/ 4<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Therefore, the ratio of the maximum height reached to the square of the flight time is&nbsp;<strong>5:4<\/strong>.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner8-309.jpeg\" alt=\"\" class=\"wp-image-248756\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>. For a projectile, the ratio of maximum height reached to the square of flight The Correct Answer and Explanation is: The correct answer is&nbsp;1) 5:4. Here is a detailed explanation of the solution. This problem requires understanding the fundamental equations of projectile motion. We need to find the ratio of the maximum height a [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-248754","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248754","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=248754"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248754\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=248754"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=248754"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=248754"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}